91Ó°ÊÓ

The study of matter and energy at the most fundamental level is known as quantum physics. Its goal is to learn more about the characteristics and behaviors of nature's fundamental building elements.

(a)

The momentum of the value \(\rho \) of the photon is written in terms of the Plank's constant, of the value \(h\), and the wavelength, of the value \({\rm{\lambda }}\):

\(\rho = \frac{h}{\lambda }\)

Solve further as:

\(\begin{aligned}{}\rho &= \frac{{6.62 \times {{10}^{ - 34}}\;{\rm{J}} \cdot {\rm{s}}}}{{0.01 \times {{10}^{ - 9}}\;{\rm{m}}}}\\ &= 6.62 \times {10^{ - 23}}{\rm{ kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - {\rm{1}}}}\end{aligned}\)

Therefore, the momentum is \(6.62 \times {10^{ - 23}}{\rm{ kg}} \cdot {\rm{m}} \cdot {{\rm{s}}^{ - {\rm{1}}}}\).

(b)

Consider the expression for the energy of photon in terms of the momentum as:

\(E = pc\)

Here the value of \(c\) is the speed of light. Then the conversion of energy from \(J\)to \({\rm{MeV}}\), we are suppose to multiply the value by: \({\rm{6}}{\rm{.24}} \times {\rm{1}}{{\rm{0}}^{{\rm{12}}}}\;\frac{{{\rm{MeV}}}}{{\rm{J}}}\).

\(\begin{aligned}{}E &= 6.62 \times {10^{ - 23}}\;\frac{{{\rm{kg \times m}}}}{{\rm{s}}} \times 3.00 \times {10^8}\;\frac{{\rm{m}}}{{\rm{s}}}\\ &= 1.99 \times {10^{ - 14}}\;{\rm{J}}\\ &= 1.99 \times {10^{ - 14}}\;{\rm{J}} \times 6.24 \times {10^{12}}{\rm{ }}\frac{{{\rm{MeV}}}}{{\rm{J}}}\\ &= 0.124\;{\rm{MeV}}\end{aligned}\)

Therefore, the energy in MeV is:\(0.124\;{\rm{MeV}}\).